第一题(2lg2+lg3)/(1+1/2lg0.36+1/3lg8) 第二题 2√3乘以6次√12乘以3次√(3/2)
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第一题(2lg2+lg3)/(1+1/2lg0.36+1/3lg8) 第二题 2√3乘以6次√12乘以3次√(3/2)
第一题(2lg2+lg3)/(1+1/2lg0.36+1/3lg8)
第二题 2√3乘以6次√12乘以3次√(3/2)
第一题(2lg2+lg3)/(1+1/2lg0.36+1/3lg8) 第二题 2√3乘以6次√12乘以3次√(3/2)
第一题 (2lg2+lg3)/(1+1/2lg0.36+1/3lg8)
= (lg4+lg3)/(1+lg0.6+lg2)
=lg12/lg12=1
第二题 2√3*(6次√12)*(3次√(3/2))
= 2√3*(3次√(2√3))*(3次√(3/2))
=[(2√3)^(4/3)]*[(3/2)^(1/3)]
=[144^(1/3)]*[(3/2)^(1/3)]
=216^(1/3)
=6
① (lg4+lg3)/(1+lg0.6+lg2) (公式algb=lg(b^a))
=lg12/(lg10+lg1.2) (公式lga+lgb=lgab)
=lg12/lg12
=1
② =3^1/2 *12^1/6 *(3/2)^1/3
=3^1/2 *3^1/6 *4^1/6 *3^1/3 *(1/2)^1/3
=3^(1/2+1/6+1/3) *4^1/6 *4^(-1/2 *1/3)
=3 *4(1/6 -1/6)
=3*1
=3
1 6
原式=(lg4+lg3)/(1+lg0.6+lg2)=lg12/lg(10×0.6×2)=1
原式=(2√3)×3次√(√12)×3次√(3/2)
=2√3×3次√(3√3)=2√3×3次√(√27)
=2√3×√(3次√27)=6